|
| Rauber, Joel wrote:
|
| > I don't mean to imply that, if it were ordinary multiplication,
| > "canceling" implies dividing both sides by C, that is the
| inverse of
| > multiplication, hitting the l.h.s of
| >
| > C*A=C*B with C^(-1)
| >
| > That's all I meant.
| >
| > Of course it isn't ordinary multiplication . . . (hence the
| > discussion)
|
| In Clifford algebra, you _can_ divide by any nonzero vector.
Absolutely!
But it should be pointed out for the non-cognoscenti that the ordinary
vector dot product is not the vector multiplication that has an inverse
in Clifford Algebras.
|
| Being able to divide by vectors is sometimes very convenient.
|
Indeed! And I hope (trust) that no one inferred from my post the
contrary.